In a radioactive decay chain, the initial nucleus is $^{232}_{90}Th$ . At the end there are $6 \alpha$-particles and $4 \beta$-particles which are emitted. If the end nucleus, If $^{A}_{Z} X$ , $A$ and $Z$ are given by :
- A = 208; Z = 80
- A = 202; Z = 80
- A = 200; Z = 81
- A = 208; Z = 82
The Correct Option is D
Solution and Explanation
The correct answer is (D) : A = 208; Z = 82
When one α−particle is emitted, then the mass number (A) of daughter nuclei decreases by 4 and the atomic number decreases by 2.
\(^{232}_{90}Th\)\(→^{208}_{78}Y+6(^4_2He)\)
When one β−particle is emitted, then the mass number (A) of daughter nuclei increases by 1 and the atomic number remains the same.
\(^{208}_{78}Y→ ^{208}_{82}X+4β\)
Therefore, for the end nucleus, A = 208 : Z = 82
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Nuclei
In the year 1911, Rutherford discovered the atomic nucleus along with his associates. It is already known that every atom is manufactured of positive charge and mass in the form of a nucleus that is concentrated at the center of the atom. More than 99.9% of the mass of an atom is located in the nucleus. Additionally, the size of the atom is of the order of 10-10 m and that of the nucleus is of the order of 10-15 m.
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Following are the terms related to nucleus:
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